Geometric Series: Finding the nth Term

[Helios]

This looks almost like a geometric series;

1, 2, 5, 14, 41, 122, 365, ...

but each term is one less than three times the preceeding one. So is this a sequence or a series? What is a formula for the value of the nth term in terms of n?

 

[disregardthat]

So basically you have

(1) \ \ \a_{n+1} = 3a_n-1 \ \ \

and a_1 = 1.


I suggest you try to expand a_n using (1) given above, hypothesise a reasonable formula for a_n and prove your suggestion by induction.

A trick however: Set b_n = a_n- \frac{1}{2}. In that case the formula (1) is reduced to b_{n+1} + \frac{1}{2} = 3(b_n + \frac{1}{2})-1 \Leftrightarrow b_{n+1}=3b_n. This you can surely solve easily.

 

[Gerenuk]

You should write out what a(n) is in terms of a(1). It's quite easy. On part will be a geometric series.

 

[HallsofIvy]

This looks almost like a geometric series;

1, 2, 5, 14, 41, 122, 365, ...

but each term is one less than three times the preceeding one. So is this a sequence or a series? What is a formula for the value of the nth term in terms of n?

Do you not understand the difference between a sequence and a series? A series is a sum of numbers. It has nothing to do with being "geometric" or not. '1, 3, 9, 27, ... is a goemetric sequence. 1+ 3+ 9+ ... is a geometric series.

What you give is a sequence because there is no sum. 1+ 2+ 5+ 14+ ... would be a series.

 

[Helios]

ok, this looks like it's it,

a ( n ) = [1 + 3^( n - 1 )] / 2 = 1, 2, 5, 14, 41, 122, 365, ...

 

[snipez90]

Well a series is a sequence if we take a series to be the limit of its partial sums, which in many scenarios is the case. But yeah I consider even basic questions like "is this a sequence or a series" to be unimportant. Basic logic indicates that it is a list of numbers, and not a sum, so who cares what it's called if you need to find the n-th term?